Question

Combustion of a 0.9827-g sample of a compound containing only carbon, hydrogen, and oxygen produced 1.900 g of CO2 and 1.070 g of H2O. What is the empirical formula of the compound

Answer 1

Answer:

The answer to your question is   C₂H₆O

Explanation:

Data

Compound   CxHyOz

mass of the sample = 0.9827

mass of CO₂ = 1.90 g

mass of H₂O = 1.070 g

Process

1.- Find the moles and grams of Carbon

                    44g of CO₂ ------------------ 12 g of C

                     1.9 g of CO₂ ---------------  x

                     x = (1.9 x 12) / 44

                     x = 0.52 g

                    12 g of C --------------------- 1 mol

                    0.52 g     -------------------- x

                    x = (0.52 x 1) / 12

                    x = 0.043 moles

2.- Find the moles and grams of Hydrogen

                   18 g of H₂O ----------------- 2 g of H

                   1.07 g of H₂O --------------  x

                   x = (1.07 x 2)/18

                   x = 0.12 g of H

                   1 g of H -------------------- 1 mol of H

                   0.12 g   --------------------- x

                   x = (0.12 x 1)/1

                   x = 0.12 moles of H

3.- Calculate the moles and grams of Oxygen

Mass of Oxygen = 0.9827 - 0.12 - 0.52

                            = 0.3427 g

                    16 g of O ----------------- 1 mol

                     0.3427 g ---------------- x

                     x = (0.3427 x 1) / 16

                     x = 0.021 moles

4.- Divide by the lowest number of moles

Carbon       0.043 / 0.021 = 2

Hydrogen   0.12 / 0.021 = 5.7 ≈ 6

Oxygen       0.021 / 0.021 = 1

5.- Write the empirical formula

                    C₂H₆O

Answer 2

Answer:

The empirical formula of the compound is C4H11O2

Explanation:

Step 1: Data given

Mass of the sample = 0.9827 grams

Mass of CO2 produced = 1.900 grams

Mass of H2O produced = 1.070 grams

Molar mass of CO2 = 44.01 g/mol

Molar mass of H2O = 18.02 g/mol

Molar mass of O2 = 32.0 g/mol

Step 2: Calculate moles CO2

Moles CO2 = mass CO2 / molar mass CO2

Moles CO2 = 1.900 grams / 44.01 g/mol

Moles CO2 = 0.04317 moles

Step 3: Calculate moles C

In 1 mol CO2 we have 1 mol C

For 0.04317 moles CO2 we have 0.04317 moles C

Step 4: Calculate mass C

Mass C = 0.04317 moles * 12.01 g/mol

Mass C = 0.5185 mass

Step 5: Calculate moles H2O

Moles H2O = 1.070 grams / 18.02 g/mol

Moles H2O = 0.05938 moles

Step 6: Calculate moles H

In 1 mol H2O we have 2 moles H

In 0.05938 moles H2O we have 2*0.05938 = 0.11876 moles H

Step 7: Calculate mass H

Mass H = 0.11876 * 1.01 g/mol

Mass H = 0.1199 grams

Step 8: Calculate mass O

Mass O = 0.9827 grams - 0.5185 grams - 0.1199 grams

Mass O = 0.3443 grams

Step 9: Calculate moles O

Moles O = 0.3443 grams / 16.0 g/mol

Moles O = 0.02152 moles

Step 10: Calculate mol ratio

We divide by the smallest amount of moles

C: 0.04317 moles / 0.02152 moles  = 2

H: 0.11876 moles / 0.02152 moles = 5.5

O: 0.02152 moles / 0.02152 moles = 1

For each 2 C atoms we have 5.5 H atoms and 1 O atom

This means for each 4 C atoms we have 11 H atoms and 2 O atoms

The empirical formula of the compound is C4H11O2